Back to the on-screen lesson ·
A second operation, distributing over the first, and the ladder it creates: ring, commutative ring, integral domain, field — with the two questions that place any example on it.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to check the ring axioms, say which rung of the ladder a given ring sits on, distinguish zero divisors from units and explain why nothing is both, compute the characteristic of a ring and of a product of rings, decide when the integers modulo $n$ form a field, and prove that a finite integral domain is always a field.
Every group in this course had one operation. But the integers have two, and so do the polynomials, and so do the residues modulo $n$: addition and multiplication, tied together by distributivity. Almost all the structure of unit 1 survives, with one striking change — multiplication does not give inverses, and that single failure is what the whole of this unit is about.
A ring is a set with two operations: an abelian group under $+$, an associative multiplication, and distributivity both ways. It is commutative when multiplication commutes and has a unity when there is a multiplicative identity $1$. A zero divisor is a non-zero $a$ with $ab = 0$ for some non-zero $b$; an integral domain is a commutative ring with $1 \ne 0$ and no zero divisors; a field is a commutative ring with $1 \ne 0$ in which every non-zero element has an inverse. The characteristic is the least $n > 0$ with $n$ copies of $1$ summing to $0$, or $0$ if there is none.
A ring $R$ has two operations. Under $+$ it is an abelian group, with identity $0$ and negatives. Multiplication is associative and distributes over addition on both sides. That is all: multiplication need not commute, there need not be a $1$, and nothing has a multiplicative inverse by right.
From there the ladder climbs by adding conditions.
| Structure | Adds |
|---|---|
| ring | — |
| commutative ring with $1$ | $ab = ba$, and a multiplicative identity |
| integral domain | no zero divisors, equivalently cancellation |
| field | every non-zero element is invertible |
The two rungs that carry the content are the last two.
No zero divisors is cancellation. In a domain, $ab = ac$ with $a \ne 0$ gives $b = c$: this is exactly the condition that made the group arithmetic of unit 1 work, now imposed by hand because multiplication supplies no inverses.
Every field is a domain, since $ab = 0$ with $a \ne 0$ gives $b = a^{-1}ab = 0$. The converse fails — the integers — except when the ring is finite, which is this lesson's last item and the reason finite fields exist at all.
A few facts hold in every ring and are worth proving once: $0a = 0$, $(-a)b = -(ab)$, and $(-a)(-b) = ab$. None is an axiom; each comes out of distributivity in two lines.
The characteristic is $0$ for the integers, the rationals and the polynomials over them, and $n$ for the integers modulo $n$. In a domain it is $0$ or prime, because a composite characteristic would factor into a pair of zero divisors.
An algebraic fact is about the operation, not about what the elements happen to be. The same group turns up as rotations of a square, as residues under addition and as matrices, and a theorem proved once from the axioms holds in all of them at once.
Another way: picture
Think of the integers and ask which familiar abilities survive in a general ring. Adding, subtracting and multiplying: always. Cancelling a common factor: only in a domain. Dividing: only in a field. The ladder is a list of the abilities you are allowed to assume, and the point of naming the rungs is that a theorem proved on a low rung holds for everything above it.
Another way: steps
To place a structure on the ladder: 1. Check it is a ring: abelian under addition, associative multiplication, distributive. 2. Does multiplication commute? Is there a $1$? 3. Can two non-zero elements multiply to zero? If so, it stops at commutative ring. 4. Does every non-zero element have an inverse? If so, it is a field. 5. If it is finite and a domain, it is automatically a field.
| Ring | Commutative | Zero divisors | Units | Rung |
|---|---|---|---|---|
| $\mathbb{Z}$ | yes | none | $\pm 1$ | domain |
| $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$ | yes | none | everything non-zero | field |
| $\mathbb{Z}_p$, $p$ prime | yes | none | everything non-zero | field |
| $\mathbb{Z}_n$, $n$ composite | yes | yes | the classes coprime to $n$ | commutative ring |
| $\mathbb{Q}[x]$ | yes | none | the non-zero constants | domain |
| $M_2(\mathbb{R})$ | no | yes | the invertible matrices | ring |
| $\mathbb{Z}_m \times \mathbb{Z}_n$ | yes | yes | pairs of units | commutative ring |
Three of those repay a second look.
$\mathbb{Q}[x]$ is a domain because degrees add: the leading coefficient of a product is the product of the leading coefficients, which is non-zero in a field. That single observation is why the whole of unit 6 works.
$M_2(\mathbb{R})$ fails in two independent ways. Multiplication does not commute, and there are zero divisors — a matrix with a zero row times one with a matching zero column. Yet it has plenty of units, the invertible matrices, so having many units does not place a ring high on the ladder.
A product ring always has zero divisors, since $(1, 0)(0, 1) = (0, 0)$, no matter how good the factors are. So $\mathbb{Z}_2 \times \mathbb{Z}_2$ is not a field even though both factors are, and this is why the finite fields of the last lesson cannot be built as products.
Assuming multiplication commutes. The matrices do not, and neither do the quaternions. If a proof uses $ab = ba$, the hypothesis commutative must be present.
Assuming a $1$ exists. The even integers form a ring with no unity. Many books build it into the definition; this course states it when it is needed.
Expecting cancellation. $3 \times 2 = 3 \times 4$ in $\mathbb{Z}_6$ with $2 \ne 4$. Cancellation is the domain condition, not a general fact.
Thinking a zero divisor can be a unit. It cannot: multiplying $ab = 0$ by $a^{-1}$ would force $b = 0$. In a finite commutative ring with $1$, every non-zero element is one or the other.
*Reading characteristic $0$ as the characteristic is zero copies. It is the convention for never returns to zero*, and rings of characteristic $0$ are the infinite ones you know best.
Commutative with $1$. Units: the classes coprime to $12$, namely $1, 5, 7, 11$ — four of them, which is $\phi(12)$.
The units.
Zero divisors: $3 \times 4 = 12 \equiv 0$ with neither factor zero, so $3$ and $4$ are zero divisors, as are $2, 6, 8, 9, 10$.
Zero divisors.
Four units, seven zero divisors and the element $0$: eleven non-zero elements, each a unit or a zero divisor, and nothing is both.
The ring stops at commutative ring.
Take non-zero $f$ and $g$ in $\mathbb{Q}[x]$, of degrees $m$ and $n$, with leading coefficients $a$ and $b$.
Two non-zero polynomials.
The coefficient of $x^{m+n}$ in $fg$ is $ab$, which is non-zero because $\mathbb{Q}$ is a field.
The top term cannot cancel.
So $fg \ne 0$: no zero divisors. Over $\mathbb{Z}_6$ the same argument fails, and indeed $2x \cdot 3x = 0$ there.
The coefficient ring has to be a domain.
Addition: an abelian group. Multiplication: associative, commutative, and distributive. So it is a commutative ring.
It is a ring.
No zero divisors, since a product of non-zero integers is non-zero.
Cancellation holds.
But there is no $1$: no even number multiplies every even number to itself. So it is a commutative ring without unity, and the words integral domain do not apply, because that definition asks for a $1$.
Match each ring to its place on the ladder from ring to field.
| an integral domain that is not a field | a field | a commutative ring with zero divisors | a ring whose multiplication does not commute | not a ring at all | |
|---|---|---|---|---|---|
| The rationals | |||||
| The integers | |||||
| The integers modulo $6$ | |||||
| The two-by-two real matrices |
The characteristic of a ring is the least number of times $1$ has to be added to itself to give $0$, or $0$ when that never happens. Fill in the characteristic of each ring.
| Its characteristic | |
|---|---|
| the integers | |
| the integers modulo 9 | |
| the rationals | |
| the polynomials with rational coefficients |
Select every statement that is true.
This task has no paper form; do it on a device.
Are the integers modulo $7$ a field?
What is the characteristic of the ring of pairs, with first coordinate taken modulo $2$ and second modulo $6$, added and multiplied coordinate by coordinate?
Answer:
Build the proof that a finite integral domain is a field.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Are the integers modulo $11$ a field?
You can place a ring on the ladder from ring to field and say which condition each rung adds. Say in your own words why cancellation is the same condition as having no zero divisors. Next: units and zero divisors counted, in the ring where every non-zero element is one or the other.
9. Your turn: is the set of even integers a ring, and where does it stop on the ladder?, step 3